Answer:
The tensions in  is approximately 4,934.2 lb and the tension in
 is approximately 4,934.2 lb and the tension in  is approximately  6,035.7 lb
 is approximately  6,035.7 lb
Explanation:
The given information are;
The angle formed by the two rope segments are;
The angle, Φ, formed by rope segment BC with the line AB extended to the center (midpoint) of the ship = 26.0°
The angle, θ, formed by rope segment BD with the line AB extended to the center (midpoint) of the ship = 21.0°
Therefore, we have;
The tension in rope segment BC = 
The tension in rope segment BD = 
The tension in rope segment AB =  = Pulling force of tugboat = 1200 lb
 = Pulling force of tugboat = 1200 lb
By resolution of forces acting along the line A_F gives;
 × cos(26.0°) +
 × cos(26.0°) +  × cos(21.0°) =
 × cos(21.0°) =  = 1200 lb
 = 1200 lb
 × cos(26.0°) +
 × cos(26.0°) +  × cos(21.0°) = 1200 lb............(1)
 × cos(21.0°) = 1200 lb............(1)
Similarly, we have for equilibrium, the sum of the forces acting perpendicular to tow cable = 0, therefore, we have;
 × sin(26.0°) +
 × sin(26.0°) +  × sin(21.0°) = 0...........................(2)
 × sin(21.0°) = 0...........................(2)
Which gives;
 × sin(26.0°) = -
 × sin(26.0°) = -  × sin(21.0°)
 × sin(21.0°)
 = -
 = -  × sin(21.0°)/(sin(26.0°))  ≈ -
 × sin(21.0°)/(sin(26.0°))  ≈ -  × 0.8175
 × 0.8175
Substituting the value of,  , in equation (1), gives;
, in equation (1), gives;
-  × 0.8175 × cos(26.0°) +
 × 0.8175 × cos(26.0°) +  × cos(21.0°) = 1200 lb
 × cos(21.0°) = 1200 lb
-  × 0.7348  +
 × 0.7348  +  ×0.9336 = 1200 lb
 ×0.9336 = 1200 lb
  ×0.1988 = 1200 lb
 ×0.1988 = 1200 lb
  ≈ 1200 lb/0.1988 = 6,035.6938 lb
 ≈ 1200 lb/0.1988 = 6,035.6938 lb
  ≈ 6,035.6938 lb
 ≈ 6,035.6938 lb
 ≈ -
 ≈ -  × 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb
 × 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb
 ≈ -4934.1733 lb
 ≈ -4934.1733 lb
From which we have;
The tensions in  ≈ -4934.2 lb and
 ≈ -4934.2 lb and   ≈ 6,035.7 lb.
 ≈ 6,035.7 lb.